= Fourier transform of a two-sided exponential
{c}
{title2=$\mathcal F(e^{-a|x|})(s)=2a/(a^2+s^2)$}
For $a>0$, use the <Fourier transform> convention $\widehat f(s)=\int_{\mathbb R}f(x)e^{-isx}\,dx$. Splitting the integral at zero gives $1/(a+is)+1/(a-is)=2a/(a^2+s^2)$. By the <Fourier inversion theorem>, the dual identity is
$$
\mathcal F\!\left(\frac1{a^2+x^2}\right)(s)=\frac\pi a e^{-a|s|}.
$$
This pair connects an <exponential function> with the kernel of a <Cauchy distribution> and produces the <Poisson kernel for the upper half-plane> by inversion of $e^{-y|s|}$.
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